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What are the asymptotes of #f(x)=-x/((x^2-8)(5x+2)) #?

Answer 1

William Abdalla

vertical asymptotes: #x = +- 2sqrt(2); x = -2/5#
horizontal asymptotes: #y = 0#

Given: #-x/((x^2-8)(5x+2))#

Factor the difference of squares in the denominator #a^2 - b^2 = (a-b)(a+b):#

#(x^2 - (sqrt(8))^2) = (x - sqrt(8))(x + sqrt(8))#

#sqrt(8) = sqrt(4*2) = 2sqrt(2)#

Therefore: #-x/((x^2-8)(5x+2)) = -x/((x-2sqrt(2))(x+2sqrt(2))(5x+2))#

Find vertical asymptotes #D(x) = 0#:

vertical asymptotes: #x = +- 2sqrt(2); x = -2/5#

Find horizontal asymptotes using Precalculus for a rational function: #(N(x))/(D(x)) = (a_nx^n + ...)/(b_mx^m + ...)#

When #n < m# horizontal asymptote is #y = 0#

Using Calculus to find horizontal asymptotes:

#lim x->oo -x/((x^2-8)(5x+2)) =#

#lim x->oo -x/(5x^3+2x^2-40x+16)#

Divide all by the largest #x# in the denominator:

#lim x->oo (-x/x^3)/((5x^3)/x^3 + (2x^2)/x^3 - (40x)/x^3 + 16/x^3)#

#lim x->oo (-1/x^2)/(5 + 2/x - 40/x^2 + 16/x^3) = 0/5 = 0#

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Answer 2

Christopher Agresta

The function f(x)=-x/((x^2-8)(5x+2)) has two vertical asymptotes at x = √8 and x = -√8, and one horizontal asymptote at y = 0.

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Answer from HIX Tutor

When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.